Let $a,$ $b,$ $c$ be positive real numbers such that $a + b + c = 1.$  Find the minimum value of
\[\frac{1}{a + 2b} + \frac{1}{b + 2c} + \frac{1}{c + 2a}.\]
Solution: By AM-HM,
\[\frac{(a + 2b) + (b + 2c) + (c + 2a)}{3} \ge \frac{3}{\frac{1}{a + 2b} + \frac{1}{b + 2c} + \frac{1}{c + 2a}},\]so
\[\frac{1}{a + 2b} + \frac{1}{b + 2c} + \frac{1}{c + 2a} \ge \frac{9}{3a + 3b + 3c} = \frac{9}{3} = 3.\]Equality occurs when $a = b = c = \frac{1}{3},$ so the minimum value is $\boxed{3}.$